The equation of hyperbola $H$ is $\dfrac {(y+4)^{2}}{16}-\dfrac {(x-4)^{2}}{81} = 1$. What are the asymptotes?
Solution: We want to rewrite the equation in terms of $y$ , so start off by moving the $y$ terms to one side: $\dfrac {(y+4)^{2}}{16} = 1 + \dfrac {(x-4)^{2}}{81}$ Multiply both sides of the equation by $16$ $(y+4)^{2} = { 16 + \dfrac{ (x-4)^{2} \cdot 16 }{81}}$ Take the square root of both sides. $\sqrt{(y+4)^{2}} = \pm \sqrt { 16 + \dfrac{ (x-4)^{2} \cdot 16 }{81}}$ $ y + 4 = \pm \sqrt { 16 + \dfrac{ (x-4)^{2} \cdot 16 }{81}}$ As $x$ approaches positive or negative infinity, the constant term in the square root matters less and less, so we can just ignore it. $y + 4 \approx \pm \sqrt {\dfrac{ (x-4)^{2} \cdot 16 }{81}}$ $y + 4 \approx \pm \left(\dfrac{4 \cdot (x - 4)}{9}\right)$ Subtract $4$ from both sides and rewrite as an equality in terms of $y$ to get the equation of the asymptotes: $y = \pm \dfrac{4}{9}(x - 4) -4$